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Roll of the die (Posted on 2005-04-21) Difficulty: 4 of 5
The object of the dice game is to be the first player to reach a score of at least 100 points.
Each player’s turn consists of repeatedly rolling a die.
After each roll, the player has two choices: roll again, or stop.
- If the player rolls 1, nothing is scored in that turn and it becomes the opponent’s turn.
- If the player rolls a number other than 1, the number is added to the player’s turn total and the player’s turn continues.
- If the player stops, the turn total (the sum of the rolls during the turn), is added to the player’s score, and it becomes the opponent’s turn.

What's your strategy?

See The Solution Submitted by Hugo    
Rating: 4.1111 (9 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
a wiser approach | Comment 9 of 23 |

I've worked some strategy before, but I think there are wiser ways.

It's clear that rolling just once and stopping is not a good approach to this game. On the other hand, rolling a lot of times before stopping seems also a losing approach. So: is there a optimal number of rolls, without any consideration of the score of other players? How many consecutive rolls in a turn give the best result on the long run? This would be the base for strategy.

To answer: for each new roll, the probability of non hitting 1 decreases, while instead the score increase. We have to optimaze the whole. 

Probability of non hitting 1 (=p) is, in one roll, 5/6; in n rolls is (5/6)^n. It's higher is we only roll once, and decreases with each new roll. The accumulated score with each new roll increases. As, if we are not hitting 1, we can only have a 2, 3, 4, 5, and 6 score; we can suppose that each roll add a medium score of 4 points. So the medium accumulated score (=s) after n rolls is 4*n.

The optimal approach would be obtained when  p*s is maximum. Values of p*s are:

n=1, p= 5/6, s=4, p*s=3,66

n=2, p=25/36, s=8, p*s=5,55

n=3, p=125/216, s=12, p*s=6.94

n=4, p=625/1296, s=16, p*s=7,71

n=5, p=3125/7776, s=20, p*s=8,03

n=6, p=15625/46656, s=24, p*s=8,03

n=7, p=78125/279936, s=28, p*s=7,81

This shows that optimal values occurs when we roll consecutively 5 or 6 times (is the same value for both). In this case we can expect that each turn fruits 8,03 points in media. It's clear that some turns would fruit 0, while others will add the whole accumulated score of the 5 or six rolls.

So, the strategy cuold be this. We can start the game with a five or six rolls turn, and go ahead that way for a while (8-10 turns). We probably will do quite well. When players are near the end, strategy will depend of scores. If we are doing fine, is better not to change strategy; if we are not winning, we can risk any more, putting ourshelves in the hands of fortune. 



Edited on April 23, 2005, 2:14 pm
  Posted by armando on 2005-04-23 08:40:01

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